# Why is it not possible to use the ideal gas law to determine if the pressure changed in this situation?

I was given this question for my homework:

A cylinder with a valve at the bottom is filled with an ideal gas. The vale is now opened and some of the gas escapes slowly. The valve is then closed, after which the piston is observed to be at a lower position. Assume that the system is in thermal equilibrium with the surroundings at all times.

Since the only forces acting on the piston at all times are 1) force from atmosphere 2) weight of piston 3) force from pressure of gas in cylinder

Since first two do not change, I concluded that the pressure inside the cylinder must also be constant.

Why is it not possible to use the ideal gas law to determine whether the pressure changed in the process?

And I'm not sure. It seems to be that it's because we don't have any numbers but frankly that seems a bad reason. Is that it? If not, please don't give me the answer, but if you could give me a hint that would be great.

I was also thinking that a given volume of any ideal gas contains the same number of moles, so when the number of moles decreases the volume must decrease by the same factor. Since the system is in thermal equilibrium the whole time and the ideal gas constant is, well, constant, we can use the ideal gas law to conclude the pressure does not change.

• It seems a silly question. Obviously the ideal gas equation of state still applies. My guess is that they mean Boyle's Law, Charles law, etc don't apply because the number of moles of gas has changed. – John Rennie Dec 1 '15 at 18:04
• I feel the same way and appreciate your camaraderie in this matter. – Zachary F Dec 1 '15 at 18:05
• You don't know how many moles of gas are contained in the cylinder. This means that P = nRT/V has two unknowns in it rather than one. – David White Jun 17 '16 at 13:50

Edit : Assume the outside is of infinite volume. Then the ideal gas law reads $P = \rho R T$, $\rho$ being the particle density. If we write the mechanical equilibrium of the piston, we get $P_{ext} S + m g = P_{int}$. $\rho$, $R$ and $T$ are constants, so the external pressure doesn't change. We then conclude that the internal pressure doesn't change either. We can make the same statement if the outside is of finite volume, taking into account the change in number of particles inside the cylinder (the total number of particles inside + outside being conserved). So I don't really understand the question you were given, I agree with the awnser of John Rennie.
• I did some mathematical reasoning to show that if $V_f=αV_i$ then $n_f=αn$ so I'm a bit skeptical. – Zachary F Dec 1 '15 at 17:38